A186526 Number T(n,k) of permutations on n elements with exactly k 3-cycles; triangle read by rows.
1, 1, 2, 4, 2, 16, 8, 80, 40, 520, 160, 40, 3640, 1120, 280, 29120, 8960, 2240, 259840, 87360, 13440, 2240, 2598400, 873600, 134400, 22400, 28582400, 9609600, 1478400, 246400, 343235200, 114329600, 19219200, 1971200, 246400, 4462057600, 1486284800, 249849600, 25625600, 3203200, 62468806400, 20807987200, 3497894400, 358758400, 44844800, 936987251200, 312344032000, 52019968000, 5829824000, 448448000, 44844800
Offset: 0
Examples
For n=4 and k=1, T(4,1)=8 since there are 8 permutations on 4 elements with 1 cycle of length 3, namely, (abc)(d), (acb)(d), (abd)(c), (adb)(c), (acd)(b), (adc)(b), (bcd)(a), and (bdc)(a). Triangle T(n,k) begins: : 1; : 1; : 2; : 4, 2; : 16, 8; : 80, 40; : 520, 160, 40; : 3640, 1120, 280; : 29120, 8960, 2240; : ...
References
- Arratia, R. and Tavaré, S. (1992). The cycle structure of random permutations. Ann. Probab. 20 1567-1591.
Links
- Alois P. Heinz, Rows n = 0..250, flattened
Programs
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Maple
seq(seq(n!*(1/3)^x/x!*sum((-1/3)^j/j!,j=0..(floor(n/3)-x)),x=0..floor(n/3)),n=0..15); # second Maple program: b:= proc(n) option remember; expand(`if`(n=0, 1, add(b(n-i)* `if`(i=3, x, 1)*binomial(n-1, i-1)*(i-1)!, i=1..n))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)): seq(T(n), n=0..15); # Alois P. Heinz, Sep 25 2016
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Mathematica
nn = 8; Range[0, nn]! CoefficientList[ Series[Exp[x^3/3 (y - 1)]/(1 - x), {x, 0, nn}], {x, y}] // Grid
Formula
T(n,k) = (n!(1/3)^k)/k!*sum((-1/3)^j/j!, j=0..(m-k)) where m=floor(n/3).
E.g.f.: exp(x^3/3*(y-1))/(1-x). - Geoffrey Critzer, Aug 26 2012.
Comments